[dan2915]

Sure. I used a somewhat roundabout method. Let A = F/k be the amplitude of the oscillation. The position will follow simple harmonic motion with equation x = L + A * cos(sqrt(k/m)*t), with x = 0 as the left end of the spring and the mass started at t = 0. So we set x = L2 and so L2-L = A * cos(sqrt(k/m)*t). Also, v = x' = - A * sqrt(k/m) * sin(sqrt(k/m)*t). So then we have v^2 = A^2 * (k/m) * sin^2(sqrt(k/m)*t) = (k/m) * (A^2- A^2 * cos^2(sqrt(k/m)*t)) = (k/m) * (A^2-(L2-L)^2) = ((F^2)/(k*m) - (k/m)*(L2-L)^2). Finally we have
v = sqrt((F^2)/(k*m) - (k/m)*(L2-L)^2).